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ME 300 Thermodynamics II Spring 2015
Exam 1 NAME:________________________________ PUID#:________________________________ Circle your section: Son Jain Lucht 8:30AM 11:30AM 2:30PM Instructions: This is a closed book/note exam. You may use the equation sheet available from the class website. You may use an approved calculator (see syllabus). You must start from the basic form of the governing equations and simplify accordingly where necessary to solve each problem. Hence, you must show all your work for full credit. Keep your eyes on your own paper. If you are caught cheating you will get a zero for the exam and your name will be turned over to the Dean of Students.
Question Total Score
1 30
2 35
3 35
Total (out of 100)
Name: __________________________
1. (a) (10 points) Calculate the total work, in kJ, for process 1-3 shown in the figure when the system consists of 2 kg of nitrogen.
No work is done during the process 2-3 since the area under process line is zero. Then the work done is equal to the area under the process line 1-2:
Wb,out =Area = P1 +P2
2m(v 2 −v1)
=(100+ 500)kPa
2(2 kg)(1.0− 0.5)m3 /kg 1 kJ
1 kPa ⋅m3
#
$%
&
'(
= 300 kJ
W = 300 kJ
1. (b) (10 points) Heat is lost through a plane wall steadily at a rate of 800 W. If the inner
and outer surface temperatures of the wall are 20°C and 5°C, respectively, and the environment temperature is 0°C, what is the rate of exergy destruction (W) within the wall? Put your answer in the box.
dSCVdt
=!QCVTb
+ !mii∑ si − !me
e∑ se + !σCV
Steady, closed system:
0 =!QCVTb
+ !σCV
so, !σCV = −
!QCVTb,1 + 273
−!QCV
Tb,2 + 273
!σCV = −−800W5+ 273K
−800W20+ 273K
!σCV = 0.1473W/K
ED = T0 !σCV = 273K(0.1473W/K) = 40.22W
ED = 40.22 W
Name: __________________________
1. (c) (10 points) A heat engine receives heat from a source at 1500 K at a rate of 600 kJ/s and rejects the waste heat to a sink at 300 K. If the power output of the engine is 400 kW, what is the thermal efficiency and second-law efficiency of this heat engine? Put your answer in the box.
η =W /Qin = 400kW / 600kW = 0.67ηrev =1−TL /TH =1−300 /1500 = 0.80ε =η /ηrev = 0.67 / 0.80 = 0.84
First law (Thermal) efficiency = η =0.67
Second law efficiency = ε =ηII =0.84
Name _____________________ Problem #2 (35 points)
Given: A feedwater heater has water with a mass flow rate of 5 kg/s flowing through it. The water enters the feedwater heater at a pressure of 5 MPa and a temperature of 40°C. The water is being heated from two sources. One source adds 900 kW from a 100°C reservoir and the other source adds heat from a 200°C reservoir so that the water exit conditions are 5 MPa and 180°C. Let T0 = 25°C and p0 = 1 atm. Find: (a) Draw the control volume boundaries such that the temperature at the heat transfer surface is
known. [2 pts] (b) State the assumptions that you are making and list the basic equations for your analysis. [3
pts] (c) Calculate the rate at which exergy is supplied to the water flowing through the heater (in
kW). [15 pts]
(d) Calculate the entropy generation rate associated with heating the water that is flowing through the feedwater heater (in kW/K). [10 pts]
(e) Develop an expression and then calculate the exergetic efficiency of the device. [5 pts] PUT YOUR ANSWERS IN BOXES PROVIDED.
100 °C200 °C
1Q!
2Q!
(a)
Assumptions: The control volume is at steady-state.
Name: __________________________ Basic equations: dmcvdt
= !mi − !me ⇒ !mi = !me
dEcvdt
= !Q − !W + !mii∑ hi +
Vi2
2+ gzi
!
"##
$
%&&− !me
e∑ he +
Ve2
2+ gze
!
"##
$
%&&
dScvdt
=!Qj
Tji=1
j
∑ + !mii∑ si − !me
e∑ se + !σ
E.
d =T0 !σ
Part (c): dmcvdt
= !mi − !me ⇒ !mi = !me
dEcvdt
= !Q − !W + !mii∑ hi +
Vi2
2+ gzi
!
"##
$
%&&−
!mee∑ he +
Ve2
2+ gze
!
"##
$
%&&
0 = !Q1+ !Q2 + !m hi − he( )!Q2 = − !Q1− !m hi − he( )
Using Table A-5, ih and eh can be evaluated.
ih is evaluated at 40iT
°= C and 5p = MPa; 171.9ih = kJ/kg.
eh is evaluated at 180eT°= C and 5p = MPa; 765.25eh = kJ/kg.
!Q2 = 900−5 171.9−765.25( ) = 2067 kW
The total exergy supplied to the water by the heaters is given by
i=1
j
∑ 1−T0
Tj
#
$%%
&
'((!Qj = 1−
T0
T1
#
$%%
&
'((!Q1 + 1−
T0
T2
#
$%%
&
'((!Q2
= 1− 25+ 273.15100+ 273.15
#
$%
&
'(900+ 1− 25+ 273.15
200+ 273.15#
$%
&
'(2067
= 945.3 kW
Part (d): dmcvdt
= !mi − !me ⇒ !mi = !me
dScvdt
=!Qj
Tji=1
j
∑ + !mii∑ si − !me
e∑ se + !σ
!σ = −!Q1
T1
−!Q2
T2
+ !m se − si( )
!σ = −900
373.15−
2067473.15
+5 2.134−0.5703( )!σ =1.039 kW/K
(c) 945.3 kW
(d) 1.039 kW/K
Name: __________________________ Part e. The exergetic (second-law) efficiency of the system is given by
ε= Exergy recoveredExergy supplied
=1− Exergy destroyedExergy supplied
ε =1−T0!σ
Exergy supplied= 1− (25+ 273.15)1.039
945.3= 0.6722
(e) ε = 0.6722
Problem #3 (35 points) Given: A piston-cylinder system (PCS) with an initial volume of 0.05 m3 contains R134a at an initial temperature of 12°C and quality of 0.1. The PCS is connected to a large supply line of R134a at a pressure of 12.0 bars and a temperature of 50°C. The valve to the supply line is opened and 5.0 kg of R134a flow into the PCS. The valve is then closed. At the final state, the PCS contains R134a in the saturated vapor state. During the filling process, the PCS exchanges heat with a temperature reservoir. There is no heat transfer between the PCS and the surroundings, only with the temperature reservoir. Find: (a) State the assumptions that you are making and list the basic equations for your analysis. (b) Calculate the final volume (m3) of the tank. (c) Calculate the heat transfer Q12 (kJ) for the process. (d) Is the temperature of the reservoir, Tres, greater than or less than 12°C? Justify your
answer.
Assumptions: (1) Quasiequlibrium process (2) uniform flow at inlet, i Lh h= (3) ΔPE = 0 (4) ΔKE = 0 Basic equations:
( )
( ) ( )
2 1
2 1 2 2 1 1 12 12 2 1
2
12 2 1 2 2 1 11v v
CVi i
CVCV CV i i L
dm m m m mdtdE Q W m h U U m u m u Q W m m hdt
W pd p p m m
= ⇒ = −
= − + ⇒ − = − = − + −
= ∀ = ∀ −∀ = −∫
&
& & &
Name: __________________________ State 1:
( ) ( )( ) ( )( )
( ) ( )( ) ( )( )
1 2 3
3 33
1 1
33 3
1 1 1 1 1
1 1
1 1 1 1 1
4.4294 442.94
v 0.7971 10 , v 0.04640
v 1 v v 0.9 0.7971 10 0.1 0.04640 5.36 10
65.83 , 233.63
1 0.9 65.83 0.1 233.63 82.6
f g
f g
f g
f g
kJp p barsm
m mkg kg
mx xkg
kJ kJu ukg kg
kJu x u x ukg
−
− −
= = =
= × =
= − + = × + = ×
= =
= − + = + =
State 2: 3
2 2 2 2v v 0.04640 , 233.63g gm kJu ukg kg
= = = =
Supply Line:
275.52LkJhkg
=
( )
( ) ( )( )
31
1 331
2 1
33
2 2 2
312 2 1 3
12 2 2 1 1 12 2 1
0.05 9.33v 5.36 10
5 14.3
v 14.3 0.04640 0.664
442.94 0.664 0.05 272
L
mm kgmkg
m m kg
mm kg mkg
kJW p m kJm
Q m u m u W m m h
−
∀= = =×
= + =
⎛ ⎞∀ = = =⎜ ⎟
⎝ ⎠
⎛ ⎞= ∀ −∀ = − =⎜ ⎟⎝ ⎠= − + − −
(b) 0.664 m3
( ) ( ) ( )12
12
14.3 233.63 9.3 82.6 272 5.0 275.52
1470
kJ kJ kJQ kg kg kJ kgkg kg kg
Q kJ
⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠= +
(d) , 12 .resHeat transfer is from the reservoir to the PCS therefore T C> °
(c) Q = +1470 kJ
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